All categories

3 years ago

PLEASE HELP ME ASAP

Mathematics

1 answer:

4vir4ik [10]3 years ago

8 0

Pretty sure it’s b, good luck

You might be interested in

How do u write 19/25 fraction or mixed number as a decimal

Romashka [77]

Divide the bottom into the top for a decimal

8 0

3 years ago

Read 2 more answers

Questions 1-6 please

MakcuM [25]

Question 1 is $6 \sqrt{2}$
question 2 is $4 \sqrt{5} + 3 \sqrt{2}$
question 3 is $6 \sqrt{2}$
question 4 is $8 \sqrt{3}$
question 5 is $\frac{ \sqrt{3}}{2}- \sqrt{3}$
question 6 is $5$

6 0

3 years ago

If the original statement is in the form "If a, then b," then the inverse is in the form "_____" answers attached

Zanzabum

Answer:

If not a then not b

Step-by-step explanation:

Probably gonna want to watch a video that goes over everything. Or do a search online for converse, inverse and contrapositive.

3 0

3 years ago

Jabri is thinking of three numbers. The greatest number is twice as large as the least number. The midde number is three more th

quester [9]

Answer:

Step-by-step explanation:

Givens

Let the least number = x

Let the largest number = 2x

Let the second number = x + 3

The sum = 75

Equation

x + 2x + x + 3 = 75 Combine the terms on the left

Solution

4x + 3 = 75 Subtract 3 from both sides

4x +3-3= 75-3

4x = 72 Divide by 4

4x/4 = 72/4

x = 18

Answer

Smallest number = 18

Second number = 18 + 3 = 21

Largest number = 2*18 = 36

Sum = 75

6 0

3 years ago

Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur

stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Three heads.
Event B: Fair coin.

The probability associated with 3 heads are:

$0.5^3 = 0.125$ out of 0.5(fair coin).
1 out of 0.5(biased).

Hence:

$P(A) = 0.125 + 0.5 = 0.625$

The probability of 3 heads and the fair coin is:

$P(A \cap B) = 0.5(0.125) = 0.0625$

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1$

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0

3 years ago